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3 years ago

Given f(x) = 2x^2 -3x and g(x) = x+4

Mathematics

2 answers:

Alenkasestr [34]3 years ago

8 0

Answer:

2x³ + 5x² - 12x

Step-by-step explanation:

The correct answer doesn't seem to be listed...

(fg)(x) = (2x² - 3x)(x + 4)

Which equals...

2x²(x) + 2x²(4) + (-3x)(x) + (-3x)(4)

Which simplifies to...

2x³ + 8x² - 3x² - 12x

2x³ + 5x² - 12x

Lelechka [254]3 years ago

3 0

Answer:

A is the answer to this question. Hope this helps

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The pharmacist uses 67mg, 100g and 0.6 kg of drug to compound 3 separate precription. how much total drug did the pharmacist use

FromTheMoon [43]

Answer: 700.067 g

Step-by-step explanation:

Given : The pharmacist uses 67mg, 100g and 0.6 kg of drug to compound 3 separate prescription.

We know that 1 gram= 1000 milligram

and 1 kilogram = 1000 grams

Then, By UNITARY METHOD, $\text{1 mg}=\dfrac{1}{1000}\text{ gram}$

Now, We convert each amount of drug into grams as ,

$67\ mg=67\times\dfrac{1}{1000}\ g\\\\=0.067\text{ g}$

AND

$0.6\ kg=0.6\times100=60\text{ g}$

Now, the total drug used by Pharmacist (in grams) = $0.067+100+600=700.067\ g$

Hence, Pharmacist used 700.067 g of total drug.

8 0

3 years ago

2x-y=4 what’s the answer

avanturin [10]

Answer:

2(2)-0=4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the average rate of change of the function f(x)=20(14)xf(x)=20(14)x from x = 1 to x = 2? Enter your answer, as a decimal

notsponge [240]

$\bf slope = m = \cfrac{rise}{run} \implies 
\cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby 
\begin{array}{llll}
average~rate\\
of~change
\end{array}\\\\
-------------------------------\\\\
f(x)= 20(14)^x \qquad 
\begin{cases}
x_1=1\\
x_2=2
\end{cases}\implies \cfrac{f(2)-f(1)}{2-1}
\\\\\\
\cfrac{[20(14)^2]~~-~~[20(14)^1]}{1}\implies \cfrac{20\cdot 196~~-~~20\cdot 14}{1}
\\\\\\
\cfrac{3920~~-~~280}{1}\implies \cfrac{3640}{1}\implies 3640$

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3 years ago

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Equivalent measures for 8 fl oz are ?

Gala2k [10]

If you are looking for gallons then its 0.0625

4 0

4 years ago

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$